integral from 0 to 1 of sqrt(1+(2y)^2)

Lösung

\int_{0}^{1} 1 + (2 y)^{2} d y

\frac{1}{2} (5 + \frac{1}{2} ln (2 + 5))

Dezimale

1.47894 \dots

Schritte zur Lösung

\int_{0}^{1} 1 + (2 y)^{2} d y

Trigonometrische Substitution anwenden

= \int_{0}^{a r c t a n (2)} \frac{1}{2} sec^{3} (u) d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{a r c t a n (2)} sec^{3} (u) d u

Wende integrale Reduktion an

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (2)} sec (u) d u)

\int_{0}^{a r c t a n (2)} sec (u) d u = ln (2 + 5)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5))

Vereinfache

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5)) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5))

Berechne die Grenzen:

5

= \frac{1}{2} (5 + \frac{1}{2} ln (2 + 5))

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